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Positive solutions for Hadamard differential systems with fractional integral conditions on an unbounded domain

  • Tariboon Jessada EMAIL logo , Sotiris K. Ntouyas , Suphawat Asawasamrit and Chanon Promsakon
Published/Copyright: May 25, 2017

Abstract

In this paper, we investigate the existence of positive solutions for Hadamard type fractional differential system with coupled nonlocal fractional integral boundary conditions on an infinite domain. Our analysis relies on Guo-Krasnoselskii’s and Leggett-Williams fixed point theorems. The obtained results are well illustrated with the aid of examples.

MSC 2010: 26A33; 34A08; 34B18; 34B40

1 Introduction

Fractional calculus and fractional differential equations have been studied extensively during the last decades. Fractional derivatives provide a more excellent tool for the description of memory and hereditary properties of various materials and processes than integer derivatives. Engineers and scientists have developed new models that involve fractional differential equations. These models have been applied successfully in, e.g., physics, biomathematics, blood flow phenomena, ecology, environmental issues, viscoelasticity, aerodynamics, electrodynamics of complex medium, electrical circuits, electron-analytical chemistry, control theory, etc. For a systematic development of the topic, we refer to the books [1]-[7]. As an important issue for the theory of fractional differential equations, the existence of solutions to kinds of boundary value problems has attracted many scholars attention, and lots of excellent results have been obtained by means of fixed point theorems, upper and lower solutions technique, and so forth. A variety of results on initial and boundary value problems of fractional differential equations and inclusions can easily be found in the literature on the topic. For some recent results, we can refer to [8]-[18] and references cited therein.

Boundary value problems on infinite intervals appear often in applied mathematics and physics. Due to the fact that an infinite interval is noncompact, the discussion about boundary value problem on the infinite intervals is more complicated. Results on the existence of solutions of boundary value problems on infinite intervals for differential, difference and integral equations may be found in the monographs [19,20]. For boundary value problems of fractional order on infinite intervals we refer to [21]-[25].

Many researchers have shown their interest in the study of systems of fractional differential equations. The motivation for those works stems from both the intensive development of the theory of fractional calculus itself and the applications. See for example [26]-[30] where systems for fractional differential equations were studied by using Banach contraction mapping principle and Schaefer’s fixed point theorem.

Recently in [31] we investigated the existence of positive solutions for fractional differential equations of Hadamard type, with integral boundary condition on infinite intervals

Dαu(t)+a(t)f(u(t))=0,1<α2,t(1,), (1)

u(1)=0,Dα1u()=i=1mλiIβiu(η),(2)

where Dα denotes the Hadamard fractional derivative of order α, η ∈ (1, ∞) and Iβi is the Hadamard fractional integral of order βi > 0, i = 1, 2,... , m and λi ≥ 0, i = 1, 2,... , m are given constants. For some recent results on positive solutions of fractional differential equations we refer to [32]-[37] and references cited therein.

In [38] the existence of positive solutions were studied for the following fractional system of differential equations subject to the nonlocal Riemann-Liouville fractional integral boundary conditions of the form

Dpx(t)+f(t,x(t)),y(t))=0,1<p2,t(0,1),Dqy(t)+g(t,x(t)),y(t))=0,1<q2,t(0,1),x(0)=0,x(1)=i=1mαiIγiy(η),y(0)=0,y(1)=j=1nβiIμjx(ξ),(3)

where DΦ are Riemann-Liouville fractional derivatives of orders ϕ{p,q},f,gC([0,1]×R+2,R+),Iϕ are Riemann-Liouville fractional integrals of order ϕ ∈ {γi, μj}, αi, βj > 0, i = 1,..., m, j = 1,..., n and the fixed constants 0 < η < ξ < 1.

In this paper we investigate the existence of positive solutions for the following fractional system of Hadamard differential equations subject to the fractional integral boundary conditions on an unbounded domain

Dpx(t)+a(t)f(x(t)),y(t))=0,1<p2,t(1,),Dqy(t)+b(t)g(x(t)),y(t))=0,1<q2,t(1,),x(1)=0,Dp1x()=i=1mλiIαiy(η),y(1)=0,Dq1y()=j=1nσjIβjx(ξ),(4)

where DΦ are Hadamard fractional derivatives of orders Φ ∈ {p, q} with lower limit 1,f,gC([1,]×R+2,R+), IΦ are Hadamard fractional integrals of order Φ ∈ {αi, βj} with lower limit 1, λi, σj > 0, i,..., m, j = 1,..., n.

Applying first the well-known Guo-Krasnoselskii’s fixed point theorem we obtain the existence of at least one positive solution. Next we prove the existence of at least three distinct nonnegative solutions by using Leggett-Williams fixed point theorem.

The rest of the paper is organized as follows: In section 2, we present some preliminaries and lemmas that will be used to prove our main results. We also obtain the corresponding Green’s function and some of its properties. The main results are formulated and proved in Section 3. Especially in Subsection 3.1 we prove the existence of at least one positive solution while, in Subsection 3.2, we prove the existence of at least three distinct nonnegative solutions.

Examples illustrating our results are presented in Section 4.

2 Background materials and preliminaries

In this section, we present some notations and definitions of Hadamard fractional calculus (see [4]) and present preliminary results needed in our proofs later.

Definition 2.1

([14]). The Hadamard fractional integral of order Φ with lower limit 1 for a function g : [1, ∞) → is defined as

IΦg(t)=1Γ(q)1tlogtsΦ1g(s)sds,Φ>0,(5)
provided the integral exists, where log(•) = loge(•).

Definition 2.2

([4]). The Hadamard derivative of fractional order Φ with lower limit 1 for a function g : [1, ∞) → ℝ ℝ is defined as

Dϕg(t)=1Γ(nϕ)tddtn1tlogtsnϕ1g(s)sds,n1<ϕ<n,(6)
where n = [Φ] + 1, [Φ] denotes the integer part of the real number Φ.

Lemma 2.3

([4, Property 2.24]). If a, α, β > 0 then

Daαlogtaβ1(x)=Γ(β)Γ(βα)logxaβα1.(7)
and
Iαlogtaβ1(x)=Γ(β)Γ(β+α)logxaβ+α1.(8)

Lemma 2.4

([4]). Let q < 0 and x ∈ C[1, ∞) ⋂ L1[1, ∞). Then the Hadamard fractional differential equation Dqx(t) = 0 has the solutions

x(t)=i=1nci(logt)qi,
and the following formula holds:
IqDqx(t)=x(t)+i=1nci(logt)qi,
where ci ∈ ℝ, i = 1, 2,...., n, and n — 1 < q < n.

Lemma 2.5

Suppose that the functions u,v ∈ C([1, ∞),ℝ+) and 1 < p, q ≤ 2. Then the following system

Dpx(t)+u(t)=0,t(1,),Dqy(t)+v(t)=0,t(1,),x(1)=0,Dp1x()=i=1mλiIαiy(η),y(1)=0,Dq1y()=j=1nσjIβjx(ξ),(9)
can be written in the equivalent integral equations of the form
x(t)=1Γ(p)1tlogtsp1u(s)dss+(logt)p1[Γ(q)Ω1u(s)dssΓ(q)Ωi=1mλiΓ(q+αi)1ηlogηsq+αi1u(s)dss+Λ1Ω1v(s)dss Λ1Ωj=1nσjΓ(p+βj)1ξlogξsp+βj1u(s)dss],(10)
and
y(t)=1Γ(q)1t(logts)q1v(s)dss+(logt)q1[Γ(p)Ω1v(s)dssΓ(p)Ωj=1nσjΓ(p+βj)1ξ(logξs)p+βj1u(s)dss+Λ2Ω1u(s)dssΛ2Ωi=1mλiΓ(q+αi)1η(logηs)q+αi1v(s)dss],(11)
where
Ω:=Γ(p)Γ(q)Λ1Λ2>0,
with
Λ1:=i=1mλiΓ(q)Γ(q+αi)(logη)q+αi1andΛ2:=j=1nσjΓ(p)Γ(p+βj)(logξ)p+βj1.

Proof

By applying the Hadamard fractional integral of orders p and q to both sides of the first two equations of system (9), respectively, we obtain

x(t)=1Γ(p)1tlogtsp1u(s)dss+c1(logt)p1+c2(logt)p2,y(t)=1Γ(q)1tlogtsq1v(s)dss+k1(logt)q1+k2(logt)q2,

where c1, c2, k1, k2 ∈ ℝ.

The conditions of (9) that x(1) = 0 and y(1) = 0 imply c2 = 0 and k2 = 0. Hence

x(t)=1Γ(p)1tlogtsp1u(s)dss+c1(logt)p1,(12)

and

y(t)=1Γ(q)1tlogtsq1v(s)dss+k1(logt)q1.(13)

From Lemma 2.3, we obtain

Dp1x(t)=1tu(s)dss+c1Γ(p)andDq1y(t)=1tv(s)dss+k1Γ(q).

By applying the Hadamard fractional integral of orders βj and αi to (12) and (13), and also substitution t = ξ and t = η, respectively, we get

Iβjx(ξ)=1Γ(p+βj)1ξlogξsp+βj1u(s)dss+c1Γ(p)Γ(p+βj)(logξ)p+βj1,

and

Iαiy(η)=1Γ(q+αi)1ηlogηsq+αi1v(s)dss+k1Γ(q)Γ(q+αi)(logη)q+αi1.

Using the conditions of (9) and solving the system of linear equations we find the constants c1 and k1 as

c1=Γ(q)Ω1u(s)dssΓ(q)Ωi=1mλiΓ(q+αi)1ηlogηsq+αi1v(s)dss+Λ1Ω1v(s)dssΛ1Ωj=1nσjΓ(p+βj)1ξlogξsp+βj1u(s)dss,

and

k1=Γ(p)Ω1v(s)dssΓ(p)Ωj=1nσjΓ(p+βj)1ξlogξsp+βj1u(s)dss+Λ2Ω1u(s)dssΛ2Ωi=1mλiΓ(q+αi)1ηlogηsq+αi1v(s)dss.

Substituting the values of c1 and k1 in (12) and (13), we deduce the integral equations (10) and (11), respectively.

The converse follows by direct computation. This completes the proof.  □

Lemma 2.6

(Green’s function). The integral equations ((10)) and ((11)), in Lemma 2.5, can be expressed in the form of Green’s functions as

x(t)=1G1(t,s)u(s)dss+1G2(t,s)v(s)dss,(14)
y(t)=1G3(t,s)v(s)dss+1G4(t,s)u(s)dss,(15)
where the Green’s functions Gi (t, s), i = 1, 2, 3, 4 are given by
G1(t,s)=gp(t,s)+Λ1Ωj=1nσj(logt)p1Γ(p+βj)gβjp(ξ,s),G2(t,s)=Γ(q)Ωi=1mλi(logt)p1Γ(q+αi)gαiq(η,s),G3(t,s)=gq(t,s)+Λ2Ωi=1mλi(logt)q1Γ(q+αi)gαiq(η,s),G4(t,s)=Γ(p)Ωj=1nσj(logt)q1Γ(p+βj)gβjp(ξ,s),
where
gϕ(t,s)=(logt)ϕ1(log(t/s))ϕ1I(ϕ),1st<,(logt)ϕ1Γ(ϕ),1ts<,(16)
and
gψϕ(ρ,s)=(logρ)ϕ+ψ1(log(ρ/s))ϕ+ψ1,1sρ<,(logρ)ϕ+ψ1,1ρs<,(17)
with Φ ∈ {p, q}, ψ ∈ {αj, βi}, ρ ∈ {ξ, η}.

Proof

By Lemma 2.5, we have

x(t)=1Γ(p)1t(logts)p1u(s)dss+(logt)p1[Γ(q)Ω1u(s)dssΓ(q)Ωi=1mλiΓ(q+αi)1η(logηs)q+αi1v(s)dss+Λ1Ω1v(s)dssΛ1Ωj=1nσjΓ(p+βj)1ξ(logξs)p+βj1u(s)dss]+1Γ(p)1(logt)p1u(s)dss(Γ(p)Γ(q)Λ1Λ2)ΩΓ(p)1(logt)p1u(s)dss=1gp(t,s)u(s)dss+Λ1Ω1(logt)p1v(s)dssΓ(q)Ωi=1mλi(logt)p1Γ(q+αi)1η(logηs)q+αj1v(s)dssΛ1Ωj=1nσj(logt)p1Γ(p+βj)1ξ(logξs)p+βj1u(s)dss+Λ1Λ2ΩΓ(p)1(logt)p1u(s)dss=1gp(t,s)u(s)dss+Γ(q)Ωi=1mλi(logt)p1Γ(q+αi)1(logη)q+αj1v(s)dssΓ(q)Ωi=1mλi(logt)p1Γ(q+αi)1η(logηs)q+αj1v(s)dssΛ1Ωj=1nσj(logt)p1Γ(p+βj)1ξ(logξs)p+βj1u(s)dss+Λ1Ωj=1nσj(logt)p1Γ(p+βj)1(logξ)p+βj1u(s)dss=1gp(t,s)u(s)dss+Λ1Ωj=1nσj(logt)p1Γ(p+βj)1gβjp(ξ,s)u(s)dss+Γ(q)Ωi=1mλi(logt)p1Γ(q+αi)1gαjq(η,s)v(s)dss=1G1(t,s)u(s)dss+1G2(t,s)v(s)dss,

which yields that (14) is satisfied. In a similar way, we have

y(t)=1Γ(q)1t(logts)q1v(s)dss+(logt)q1[Γ(p)Ω1v(s)dssΓ(p)Ωj=1nσjΓ(p+βj)1ξ(logξs)p+βj1u(s)dss+Λ2Ω1u(s)dssΛ2Ωi=1mλiΓ(q+αi)1η(logηs)q+αi1v(s)dss]+1Γ(q)1(logt)q1v(s)dss(Γ(p)Γ(q)Λ1Λ2)ΩΓ(q)1(logt)q1v(s)dss=1gq(t,s)v(s)dss+Λ2Ω1(logt)q1u(s)dssΓ(p)Ωj=1nσj(logt)q1Γ(p+βj)1ξ(logξs)p+βj1u(s)dssΛ2Ωi=1mλi(logt)q1Γ(q+αi)1η(logηs)q+αi1v(s)dss+Λ1Λ2ΩΓ(q)1(logt)q1v(s)dss=1gq(t,s)v(s)dss+Γ(p)Ωj=1nσj(logt)q1Γ(p+βj)1(logξ)p+βj1u(s)dssΓ(p)Ωj=1nσj(logt)q1Γ(p+βj)1ξ(logξs)p+βj1u(s)dssΛ2Ωi=1mλi(logt)q1Γ(q+αi)1η(logηs)q+αi1v(s)dss+Λ2Ωi=1mλi(logt)q1Γ(q+αi)1(logη)q+αi1v(s)dss=1gq(t,s)v(s)dss+Λ2Ωi=1mλi(logt)q1Γ(q+αi)1gαiq(η,s)v(s)dss+Γ(p)Ωj=1nσj(logt)q1Γ(p+βj)1ξgβjp(ξ,s)u(s)dss=1G3(t,s)v(s)dss+1G4(t,s)u(s)dss,

which proves that (15) holds. This completes the proof.  □

Before establishing some properties of the Green’s functions, we set

M1=1Γ(p)+Λ1Ωj=1nσj(logξ)p+Bj1Γ(p+βj),M2=Γ(q)Ωi=1mλi(logη)q+αi1Γ(q+αi),M3=1Γ(q)+Λ2Ωi=1mλi(logη)q+αi1Γ(q+αi),M4=Γ(p)Ωj=1nσj(logξ)p+Bj˙1Γ(p+βj).

Lemma 2.7

The Green’s functions Gi(t, s), i = 1, 2, 3, 4, satisfy the following properties:

  • (P1)Gi (t, s), i = 1, 2, 3, 4 are continuous for (t, s) ∈ [1, ∞) × [1,∞);

  • (P2)Gi(t, s) ≥ 0, i = 1, 2, 3, 4 for all (t, s) ∈ [1, ∞) × [1,∞);

  • (P3)G1(t,s)1+(logt)p1+(logt)q1M1,G2(t,s)1+(logt)p1+(logt)q1M2,G3(t,s)1+(logt)p1+(logt)q1M3,G4(t,s)1+(logt)p1+(logt)q1M4;
  • (P4)infξtξk1G1(t,s)1+(logt)p1+(logt)q1Λ1Ωj=1nσj(logξ)p1gβjp.(ξ,s)Γ(p+βj)(1+(logξ)p1+(logξ)q1),infξtξk1G2(t,s)1+(logt)p1+(logt)q1Γ(q)Ωj=1mλi(logη)p1gαiq(η,s)Γ(q+αi)(1+(logη)p1+(logη)q1),infξtξk1G3(t,s)1+(logt)p1+(logt)q1Λ2Ωj=1mλi(logη)q1gαiq(η,s)Γ(q+αi)(1+(logη)p1+(logη)q1),infξtξk1G4(t,s)1+(logt)p1+(logt)q1Γ(p)Ωj=1nσj(logξ)q1gβjp.(ξ,s)Γ(p+βj)(1+(logξ)p1+(logξ)q1),foranyκ1,κ2>1.

Proof

It is easy to prove that (P1) and (P2) hold.

To prove (P3), for (s, t) ∈ [1, ∞) × [1, ∞), we have

G1(t,s)1+(logt)p1+(logt)q1=gp(t,s)1+(logt)p1+(logt)q1+Λ1Ωj=1nσj(logt)p1gβjp(ξ,s)Γ(p+βj)(1+(logt)p1+(logt)q1)(logt)p1Γ(p)(1+(logt)p1+(logt)q1)+Λ1Ωj=1nσj(logt)p1gβjp(ξ,s)Γ(p+βj)(1+(logt)p1+(logt)q1)1Γ(p)+Λ1Ωj=1nσjgβjp(ξ,s)Γ(p+βj).

In similar manner, we can prove that

G 2 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 < _ Γ ( q ) Ω i = 1 n λ i g α i q ( η , s ) Γ ( q + α i ) , G 3 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 < _ 1 Γ ( q ) + Λ 2 Ω i = 1 m λ i g α i q ( η , s ) Γ ( q + α i ) , G 4 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 < _ Γ ( p ) Ω j = 1 n σ j g β j p ( ξ , s ) Γ ( p + β j ) .

From (17), it follows that (P3) holds.

To prove (P4), from the positivity of gp(t, s) and gβjp(ξ,s), j = 1,..., n, we have for any κ1 > 1,

min ξ t ξ κ 1 G 1 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 = min ξ t ξ κ 1 [ g p ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 + Λ 1 Ω j = 1 n σ j ( log t ) p 1 g β j p . ( ξ , s ) Γ ( p + β j ) ( 1 + ( log t ) p 1 + ( log t ) q 1 ) ] min ξ t ξ κ 1 [ Λ 1 Ω j = 1 n σ j ( log t ) p 1 g β j p . ( ξ , s ) Γ ( p + β j ) ( 1 + ( log t ) p 1 + ( log t ) q 1 ) ] Λ 1 Ω j = 1 n σ j ( log ξ ) p 1 g β j p . ( ξ , s ) Γ ( p + β j ) ( 1 + ( log ξ ) p 1 + ( log ξ ) q 1 ) .

In the same way, for any κ2 > 1, we have

min η t η κ 2 G 2 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 Γ ( q ) Ω i = 1 m λ i ( log η ) p 1 g α i q ( η , s ) Γ ( q + α i ) ( 1 + ( log η ) p 1 + ( log η ) q 1 ) , min η t η κ 2 G 3 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 Λ 2 Ω i = 1 m λ i ( log η ) q 1 g α i q ( η , s ) Γ ( q + α i ) ( 1 + ( log η ) p 1 + ( log η ) q 1 ) , min η t η κ 1 G 4 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 Γ ( p ) Ω j = 1 n σ j ( log ξ ) q 1 g β j p . ( ξ , s ) Γ ( p + β j ) ( 1 + ( log ξ ) p 1 + ( log ξ ) q 1 ) .

Therefore, (P4) is proved.  □

3 Main results

Define the set

E={(x,y)C([1,),R)×C([1,),R):supt[1,)|x(t)|+|y(t)|1+(logt)p1+(logt)q1<},

and the norm

(x,y)=supt[1,)|x(t)|+|y(t)|1+(logt)p1+(logt)q1.

It is clear that (E, || • ||) is a Banach space.

Lemma 3.1

Let U ⊂ E be a bounded set. If the following conditions hold:

  1. for any (u,v)(t)U,u(t)+v(t)1+(logt)p1+(logt)q1is equicontinuous on any compact interval of [1, ∞);

  2. for any ε > 0, there exists a constant T = T(ε) > 0 such that

    u(t1)+v(t1)1+(logt1)p1+(logt1)q1u(t2)+v(t2)1+(logt2)p1+(logt2)q1<ϵ(18)
    for any t1,t2 ≥ T and (u,v) ∈ U,

    then U is relatively compact in E.

The proof is similar to that of Lemma 2.8 in [31], and is omitted.

Now, we define the positive cone PE by

P={(x,y)E:x(t)0,y(t)0on[1,)},

and the operator 𝒯 : PE by 𝒯(x, y)(t) = (A(x, y)(t), B(x, y)(t)) for all t ∈ [1, ∞), where the operators A : PE and B : PE are defined by

A(x,y)(t):=1G1(t,s)a(s)f(x(s),y(s))dss+1G2(t,s)b(s)g(x(s),y(s))dss,B(x,y)(t):=1G3(t,s)b(s)g(x(s),y(s))dss+1G4(t,s)a(s)f(x(s),y(s))dss.(19)

Throughout this paper we assume that the following conditions hold:

  • (H1)The functions f, gC([0,∞) × [0, ∞), [0, ∞)), f(x, y), g(x, y) ≠ 0 on any subinterval of (0, ∞) × (0, ∞) and f, g are bounded on [0, ∞) × [0, ∞);

  • (H2)a, b : [1, ∞) → [1, ∞) are not identical zero on any closed subinterval of [1, ∞) and

    0<1a(s)dss<,0<1b(s)dss<.

Next, we are going to prove that the operator 𝒯 is completely continuous.

Lemma 3.2

Let (H1) and (H2) hold. Then 𝒯 : P → P is completely continuous.

Proof

Firstly, we will show that 𝒯 is uniformly bounded on P. Let Ω* = {(x, y) ∈ P : ||(x, y)|| ≤ r} ≤ P be a bounded set. From (H1), there exist two positive constants L1 and L2 such that

|f(x(t),y(t))|L1,|g(x(t),y(t))|L2,(x,y)Ω.

Then, from (H2), for any (x, y) ∈ Ω*, we have

T(x,y)=supt[1,)11+(logt)p1+(logt)q1[1G1(t,s)a(s)f(x(s),y(s))dss+1G2(t,s)b(s)g(x(s),y(s))dss+1G3(t,s)b(s)g(x(s),y(s))dss+1G4(t,s)a(s)f(x(s),y(s))dss]M11a(s)f(x(s),y(s))dss+M21b(s)g(x(s),y(s))dss+M31b(s)g(x(s),y(s))dss+M41a(s)f(x(s),y(s))dss(M1+M4)L11a(s)dss+(M2+M3)L21b(s)dss<.

This means that the operator 𝒯 is uniformly bounded.

Secondly, we will show that 𝒯 is equicontinuous on any compact interval of [1, ∞). For any S > 1, t1, t2 ∈ [1, S], and (x, y) ∈ Ω*, without loss of generality, we assume that t1 < t2. Then, we have

|A(x,y)(t2)1+(logt2)p1+(logt2)q1A(x,y)(t1)1+(logt1)p1+(logt1)q1|=|1G1(t2,s)1+(logt2)p1+(logt2)q1a(s)f(x(s),y(s))dss+1G2(t2,s)1+(logt2)p1+(logt2)q1b(s)g(x(s),y(s))dss1G1(t1,s)1+(logt1)p1+(logt1)q1a(s)f(x(s),y(s))dss1G2(t1,s)1+(logt1)p1+(logt1)q1b(s)g(x(s),y(s))dss||1(G1(t2,s)1+(logt2)p1+(logt2)q1G1(t1,s)1+(logt1)p1+(logt1)q1)a(s)f(x(s),y(s))dss|+|1(G2(t2,s)1+(logt2)p1+(logt2)q1G2(t1,s)1+(logt1)p1+(logt1)q1)b(s)g(x(s),y(s))dss|.

Since

| G 1 ( t 2 , s ) 1 + ( log t 2 ) p 1 + ( log t 2 ) q 1 G 1 ( t 1 , s ) 1 + ( log t 1 ) p 1 + ( log t 1 ) q 1 | | g p ( t 2 , s ) g p ( t 1 , s ) 1 + ( log t 2 ) p 1 + ( log t 2 ) q 1 + | ( log t 2 ) p 1 ( log t 1 ) p 1 + ( log t 2 ) q l ( log t 1 ) q l | g p ( t 1 , s ) ( 1 + ( log t 2 ) p 1 + ( log t 2 ) q 1 ) ( 1 + ( log t 1 ) p 1 + ( log t 1 ) q 1 ) + | ( log t 2 ) p 1 ( log t 1 ) p 1 + ( log t 2 ) p 1 ( log t 1 ) q 1 ( log t 1 ) p 1 ( log t 2 ) q 1 | ( 1 + ( log t 2 ) p 1 + ( log t 2 ) q 1 ) ( 1 + ( log t 1 ) p 1 + ( log t 1 ) q 1 ) Λ 1 Ω j = 1 n σ j g β j p ( ξ , s ) Γ ( p + β j ) ,

and

1 | g p ( t 2 , s ) g p ( t l , s ) | 1 + ( log t 2 ) p 1 + ( log t 2 ) q 1 a ( s ) f ( x ( s ) , y ( s ) ) d s s 1 t 1 | ( log t 2 ) p 1 ( log t 1 ) p 1 | + | ( log ( t 2 / s ) ) p 1 ( log ( t 1 / s ) ) p 1 | Γ ) ( 1 + ( log t 2 ) p 1 + ( log t 2 ) q 1 ) a ( s ) f ( x ( s ) , y ( s ) ) d s s + t 1 t 2 | ( log t 2 ) p 1 ( log t 1 ) p 1 | + | ( log ( t 2 / s ) ) p 1 | Γ ( p ) ( 1 + ( log t 2 ) p 1 + ( log t 2 ) q 1 ) a ( s ) f ( x ( s ) , y ( s ) ) d s s + t 2 | ( log t 2 ) p 1 ( log t 1 ) p 1 | Γ ( p ) ( 1 + ( log t 2 ) p 1 + ( log t 2 ) q 1 ) a ( s ) f ( x ( s ) , y ( s ) ) d s s ,

we have that

1(G1(t2,s)1+(logt2)p1+(logt2)q1G1(t1,s)1+(logt1)p1+(logt1)q1)a(s)f(x(s),y(s))dss0,

uniformly as t1t2. In addition, we can find that

G2(t2,s)1+(logt2)p1+(logt2)q1G2(t1,s)1+(logt1)p1+(logt1)q1(|(logt2)p1(logt1)p1|+|(logt2)p1(logt1)q1(logt1)p1(logt2)q1|)(1+(logt2)p1+(logt2)q1)(1+(logt1)p1+(logt1)q1)×Γ(q)Ωi=1mλigαiq(η,s)Γ(q+αi),

which leads to

1(G2(t2,s)1+(logt2)p1+(logt2)q1G2(t1,s)1+(logt1)p1+(logt1)q1)b(s)g(x(s),y(s))dss0,

uniformly as t1t2. Hence

A(x,y)(t2)1+(logt2)p1+(logt2)q1A(x,y)(t1)1+(logt1)p1+(logt1)q10,uniformlyast1,t2.

In the similar way, we can prove that

B(x,y)(t2)1+(logt2)p1+(logt2)q1B(x,y)(t1)1+(logt1)p1+(logt1)q10,

as t1t2 independently of (x, y) ∈ Ω*. Therefore 𝒯Ω* is equicontinuous on [1, ∞).

Thirdly, we will show that the operator 𝒯 is equiconvergent at ∞. From the first step, for any (x, y) ∈ Ω*, we have

1a(s)f(x(s),y(s))dssL11a(s)dss<,

and

1b(s)g(x(s),y(s))dssL21b(s)dss<.

Then

lim t | A ( x , y ) ( t ) 1 + ( log t ) p 1 + ( log t ) q 1 | = lim t | 1 G 1 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 a ( s ) f ( x ( s ) , y ( s ) ) d s s + 1 G 2 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 b ( s ) g ( x ( s ) , y ( s ) ) d s s | lim t 1 ( ( log t ) p 1 Γ ( p ) ( 1 + ( log t ) p 1 + ( log t ) q 1 ) + Λ 1 Ω j = 1 n σ j ( log t ) p 1 g β j p . ( ξ , s ) Γ ( p + β j ) ( 1 + ( log t ) p 1 + ( log t ) q 1 ) ) a ( s ) f ( x ( s ) , y ( s ) ) d s s + lim t 1 ( Γ ( q ) Ω i = 1 m λ i ( log t ) p 1 g α i q ( η , s ) Γ ( q + α i ) ( 1 + ( log t ) p 1 + ( log t ) q 1 ) ) b ( s ) g ( x ( s ) , y ( s ) ) d s s M 1 L 1 1 a ( s ) d s s + M 2 L 2 1 b ( s ) d s s < ,

and

limt|B(x,y)(t)1+(logt)p1+(logt)q1|=limt|1G3(t,s)1+(logt)p1+(logt)q1b(s)g(x(s),y(s))dss+1G4(t,s)1+(logt)p1+(logt)q1a(s)f(x(s),y(s))dss|limt1((logt)q1Γ(q)(1+(logt)p1+(logt)q1)+Λ2Ωi=1mλi(logt)q1gαiq(η,s)Γ(q+αi)(1+(logt)p1+(logt)q1))b(s)g(x(s),y(s))dss+limt1(Γ(p)Ωj=1nσj(logt)q1gβjp.(ξ,s)Γ(p+βj)(1+(logt)p1+(logt)q1))a(s)f(x(s),y(s))dssM3L21b(s)dss+M4L11a(s)dss<,

which imply

limtT(x,y)(t)1+(logt)p1+(logt)q1<.

Hence 𝒯Ω* is equiconvergent at infinity.

Finally, we will prove that the operator 𝒯 is continuous. Let (xn, yn)→ ∞ as n → ∞ in P. By applying the Lebesgue dominated convergence and the continuity of f and g guarantee that

1a(s)f(xn(s),yn(s))dss1a(s)f(x(s),y(s))dssasn,

and

1b(s)g(xn(s),yn(s))dss1b(s)g(x(s),y(s))dssasn.

Therefore, we get

T ( x n , y n ) T ( x , y ) = ( A ( x n , y n ) ( t ) A ( x , y ) ( t ) , B ( x n , y n ) ( t ) B ( x , y ) ( t ) ) = sup t [ 1 , ) | A ( x n , y n ) ( t ) A ( x , y ) ( t ) | + | B ( x n y n ) ( t ) B ( x , y ) ( t ) | 1 + ( log t ) p 1 + ( 1 o g t ) q 1 M 1 1 a ( s ) | f ( x n ( s ) , y n ( s ) ) f ( x ( s ) , y ( s ) ) | d s s + M 2 1 b ( s ) | g ( x n ( s ) , y n ( s ) ) g ( x ( s ) , y ( s ) ) | d s s + M 3 1 b ( s ) | g ( x n ( s ) , y n ( s ) ) g ( x ( s ) , y ( s ) ) | d s s + M 4 1 a ( s ) | f ( x n ( s ) , y n ( s ) ) f ( x ( s ) , y ( s ) ) | d s s 0 a s n .

Thus, the operator 𝒯 is continuous.

By applying Lemma 3.1, we deduce that the operator 𝒯 : PP is completely continuous. This completes the proof.  □

To prove our main results, we set the following constants

m1=Λ1Ωj=1nσj(logξ)2p+Bj2Γ(p+βj)(1+(logξ)p1+(logξ)q1),m2=Γ(q)Ωi=1mλi(logη)p+q+αi2Γ(q+αi)(1+(logη)p1+(logη)q1),m3=Λ2Ωi=1mλi(logη)2q+αi2Γ(q+αi)(1+(logη)p1+(logη)q1),m4=Γ(p)Ωj=1nσj(logξ)p+q+Bj2Γ(p+βj)(1+(logξ)p1+(logξ)q1),n1=1a(s)dss,n2=1b(s)dss,n3=ξk1ξa(s)dss,n4=ηk2ηb(s)dss,Λ3=n3(m1+m4),Λ4=n4(m2+m3),Λ5=n1(M1+M4),Λ6=n2(M2+M3).

3.1 Existence result via Guo-Krasnoselskii’s fixed point theorem

In this subsection, the existence theorems of at least one positive solution will be established using the Guo-Krasnoselskii fixed point theorem.

Theorem 3.3

(Guo-Krasnoselskii fixed point theorem) [39] Let E be a Banach space, and let 𝒫E be a cone. Assume that Φ1, Φ2 are bounded open subsets of E with 0 ∈ ϕ1, ̄ϕ1 ⊂ Φ2, and let Q : 𝒫 ∩ (̄Φ2 \ Φ1) → 𝒫 be a completely continuous operator such that:

  1. ||Qx||≥||x||, x𝒫 ∩ ∂ Φ1, and ||Qx||≤||x||, x𝒫Φ2; or

  2. ||Qx||≤||x||, x𝒫 ∩ ∂ Φ1, and ||Qx||≥||x||, x𝒫Φ2.

Then Q has a fixed point in 𝒫 ∩ (̄Φ21).

Theorem 3.4

let f, g : + × ++ be continuous functions. Suppose that there exist positive constants ρ1<ρ2,andθ1(Λ31,),θ2(Λ41,),θ3(0,Λ51)andθ4(0,Λ61). In addition, assume that the following conditions hold: (H3)f(x,y)θ1ρ12for(x,y)[0,ρ1]×[0,ρ1]andg(x,y)θ2ρ12for(x,y)[0,ρ1]×[0,ρ1];(H4)f(x,y)θ3ρ22for(x,y)[0,ρ2]×[0,ρ2]andg(x,y)θ4ρ22(x,y)[0,ρ2]×[0,ρ2].Then, the problem (4) has at least one positive solution (x, y) such that

ρ1<(x,y)<ρ2.

Proof

It follows from Lemma 3.2 that the operator 𝒯 : 𝒫𝒫 is completely continuous. Define Φ1 = {(x, y) ∈ E : ||(x, y)|| < ρ1}. Hence, for any (x, y) ∈ 𝒫Φ1, we have 0 ≤ x(t) ≤ ρ1 and 0 ≤ y(t) ≤ ρ1 for all t ∈ [1, ∞). From assumption (H3) and Lemma 2.7, we get

T ( x , y ) = sup t [ 1 , ) | A ( x , y ) ( t ) | + | B ( x , y ) ( t ) | 1 + ( log t ) p 1 + ( log t ) q 1 = sup t [ 1 , ) 1 1 + ( log t ) p 1 + ( log t ) q 1 [ 1 G 1 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s + 1 G 2 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 G 3 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 G 4 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s ] 1 inf t [ ξ , k 1 ξ ] G 1 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s + 1 inf t [ η , k 2 η ] G 2 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s + 1 inf t [ η , k 2 η ] G 3 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s + 1 inf t [ ξ , k 1 ξ ] G 4 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s Λ 1 Ω j = 1 n σ j ( log ξ ) p 1 Γ ( p + β j ) ( 1 + ( log ξ ) p 1 + ( log ξ ) q 1 ) ξ k 1 ξ g β j p ( ξ , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s + Γ ( q ) Ω i = 1 m λ i ( log η ) p 1 Γ ( q + α i ) ( 1 + ( log η ) p 1 + ( log η ) q 1 ) η k 2 η g α i q ( η , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + Λ 2 Ω i = 1 λ i ( log η ) q 1 Γ ( q + α i ) ( 1 + ( log η ) p 1 + ( log η ) q 1 ) m η k 2 η g α i q ( η , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + r τ ( p ) Ω j = 1 n σ j ( log ξ ) q 1 Γ ( p + β j ) ( 1 + ( log ξ ) p 1 + ( log ξ ) q 1 ) ξ k 1 ξ g β j p ( ξ , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s 1 2 m 1 ρ 1 θ 1 ξ k 1 ξ a ( s ) d s s + 1 2 m 2 ρ 1 θ 2 η k 2 η b ( s ) d s s + 1 2 m 3 ρ 1 θ 1 η k 2 η b ( s ) d s s + 1 2 m 4 ρ 1 θ 2 ξ k 1 ξ a ( s ) d s s = 1 2 Λ 3 θ 1 ρ 1 + 1 2 Λ 4 θ 2 ρ 1 ρ 1 ,

which implies that ||𝒯(x,y)|| ≥ ||(x,y)|| for (x,y) ∈ 𝒫 ∩ ∂Φ1.

Next, we define Φ2 = {(x,y) ∈ E : ||(x,y)|| < ρ2}. Hence, for any (x,y) ∈ 𝒫 ∩∂Φ2, we have 0 ≤ x(t) ≤ ρ2 and 0 ≤ y(t) ≤ ρ2 for all t ∈ [1,∞). Using the condition (H4), we obtain

T ( x , y ) = sup t [ 1 , ) 1 1 + ( log t ) p 1 + ( log t ) q 1 [ 1 G 1 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s + 1 G 2 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 G 3 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 G 4 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s ] 1 2 M 1 θ 3 ρ 2 1 a ( s ) d s s + 1 2 M 2 θ 4 ρ 2 1 b ( s ) d s s + 1 2 M 3 θ 4 ρ 2 1 b ( s ) d s s + 1 2 M 4 θ 3 ρ 2 1 a ( s ) d s s = 1 2 Λ 5 θ 3 ρ 2 + 1 2 Λ 6 θ 4 ρ 2 ρ 2 ,

which leads to ||𝒯(x,y)|| ≤ ||(x,y)|| for (x,y) ∈ 𝒫 ∩ ∂Φ2.

Therefore, by the first part of Theorem 3.3, we deduce that the operator 𝒯 has a fixed point in 𝒫 ∩ (̄Φ21) which is a positive solution of problem (4). Thus the problem (4) has at least one positive solution (x, y) such that

ρ1<(x,y)<ρ2.

This completes the proof. □

Similarly to the previous theorem, we can prove the following result.

Theorem 3.5

Let f, g : ℝ+ × ℝ+ → ℝ+ be continuous functions. Assume that there exist positive constants ρ1<ρ2,andθ1(Λ31,),θ2(Λ41,),θ3(0,Λ51)andθ4(0,Λ61). In addition, assume that the following conditions hold:

(H5)f(x,y)θ3ρ12for(x,y)[0,ρ1]×[0,ρ1]andg(x,y)θ4ρ12for(x,y)[0,ρ1]×[0,ρ1];

(H6)f(x,y)θ1ρ22for(x,y)[0,ρ2]×[0,ρ2]andg(x,y)θ2ρ22for(x,y)[0,ρ2]×[0,ρ2].

Then, the problem (4) has at least one positive solution (x, y) such that

ρ1<(x,y)<ρ2.

3.2 Existence result via Leggett-Williams fixed point theorem

In this subsection, the existence of at least three positive solutions will be proved using the Leggett-Williams fixed point theorem.

Definition 3.6

A continuous mapping ω : 𝒫 → [0,∞) is said to be a nonnegative continuous concave functional on the cone 𝒫 of a real Banach space E provided that

ω(λx+(1λ)y)λω(x)+(1λ)ω(y)
for all x,y𝒫 and λ ∈ [0,1].

Let a, b,d > 0 be given constants and define Pd={(x,y)P:(x,y)<d},P¯d={(x,y)P:(x,y)d}andP(ω,a,b)={(x,y)P:ω((x,y))a,(x,y)b}.

Theorem 3.7

(Leggett-Williams fixed point theorem) [40] Let 𝒫 be a cone in the real Banach space E and c>0 be a constant. Assume that there exists a concave nonnegative continuous functional ω on 𝒫 with ω (x) ≤ ||x|| for all x ∈ ̄𝒫c. Let Q : ̄𝒫c → ̄𝒫c be a completely continuous operator. Suppose that there exist constants 0 < a < b < dc such that the following conditions hold:

  1. {xP(ω,b,d):ω(x)>b}andω(Qx)>bforxP(ω,b,d);

  2. Qx<aforxa;

  3. ω(Qx)>bforxP(ω,b,c)withQx>d.

Then Q has at least three fixed points x1, x2 and x3 in ̄𝒫c. In addition, ||x1|| < a, ω(x2) > b, ||x3|| > a with ω(x3) < b.

Theorem 3.8

Let f, g : ℝ+ × ℝ+ → ℝ+ be continuous functions. Suppose that there exist two constants k1, k2 > 1 such that [ξ, k1 ξ] ∩ [η, k2η] ≠ θ. In addition, assume that there exist positive constants a < b < c satisfying

(H7)f(x,y)<a2Λ5andg(x,y)>a2Λ6for(x,y)[0,a]×[0,a];

(H8)f(x,y)<b2Λ3andg(x,y)>b2Λ4for(x,y)[b,c]×[b,c];

(H9)f(x,y)<c2Λ5andg(x,y)>b2Λ6for(x,y)[0,c]×[0,c].

Then, the problem (4) has at least three positive solutions (x1, y1), (x2, y2) and (x3, y3) such that ( x 1 , y 1 ) < a , i n f τ 1 t τ 2 ( x 2 , y 2 ) ( t ) > b a n d ( x 3 , y 3 ) > a w i t h i n f τ 1 t τ 2 ( x 3 , y 3 ) ( t ) < b , w h e r e τ 1 = max { ξ , η } , τ 2 = min { k 1 ξ , k 2 η } .

Proof

At first, we will show that the operator 𝒯 : ̄𝒫c → ̄𝒫c. For any (x, y) ∈ ̄𝒫c, we have ||(x, y)|| ≤ c. Using the condition (H9) and Lemma 2.7, we obtain

T ( x , y ) = sup t [ 1 , ) 1 1 + ( log t ) p 1 + ( log t ) q 1 [ 1 G 1 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s + 1 G 2 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 G 3 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 G 4 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s ] 1 sup t [ 1 , ) G 1 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 a ( s ) f ( x ( s ) , y ( s ) ) d s s + 1 sup t [ 1 , ) G 2 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 sup t [ 1 , ) G 3 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 sup t [ 1 , ) G 4 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 a ( s ) f ( x ( s ) , y ( s ) ) d s s c 2 Λ 5 M 1 n 1 + c 2 Λ 6 M 2 n 2 + c 2 Λ 6 M 3 n 2 + c 2 Λ 5 M 4 n 1 = c ,

which yields 𝒯 : ̄𝒫c → ̄𝒫c. Secondly, we let (x, y) ∈ ̄Pa. By applying condition (H1 ), we have

T ( x , y ) 1 sup t [ 1 , ) G 1 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 a ( s ) f ( x ( s ) , y ( s ) ) d s s + 1 sup t [ 1 , ) G 2 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 sup t [ 1 , ) G 3 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 sup t [ 1 , ) G 4 ( t , s ) 1 + ( log t ) p 1 + ( log t ) q 1 a ( s ) f ( x ( s ) , y ( s ) ) d s s < a 2 Λ 5 M 1 n 1 + a 2 Λ 6 M 2 n 2 + a 2 Λ 6 M 3 n 2 + a 2 Λ 5 M 4 n 1 = a .

This means that the condition (i i) of Theorem 3.7 is satisfied.

Thirdly, we define τ1=max{ξ,η},τ2=min{k1ξ,k2η} and also a concave nonnegative continuous functional ω on E by

ω((x,y))=infτ1tτ2|x(t)|+|y(t)|1+(logt)p1+(logt)q1.

By choosing

(x,y)(t)=((b+c)2(1+(logt)p1+(logt)q1),(b+c)2(1+(logt)p1+(logt)q1)),

we deduce that (x, y)(t) ∈ ̄𝒫(ω, b, c) and ω((x, y)) > b. Hence it follows that {(x, y) ∈ ̄𝒫(ω, b, c) : ω((x, y)) > b} ≠ Ø. Therefore, if (x, y) ∈ ̄𝒫(ω, b, c), then we have bx(t) ≤ c and by(t) ≤ c for t ∈ [τ1, τ2]. Using condition (H8) and Lemma 2.7, we obtain

ω ( T ( x , y ) ( t ) ) = inf τ 1 t τ 2 | A ( x , y ) ( t ) | + | B ( x , y ) ( t ) | 1 + ( log t ) p 1 + ( log t ) q 1 = inf τ 1 t τ 2 1 1 + ( log t ) p 1 + ( log t ) q 1 [ 1 G 1 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s + 1 G 2 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 G 3 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 G 4 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s ] 1 inf τ 1 t τ 2 G 1 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s + 1 inf τ 1 t τ 2 G 2 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s + 1 inf τ 1 t τ 2 G 3 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s + 1 inf τ 1 t τ 2 G 4 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s > b 2 Λ 3 m 1 n 3 + b 2 Λ 4 m 2 n 4 + b 2 Λ 4 m 3 n 4 + b 2 Λ 3 m 4 n 3 = b .

Therefore, ω(𝒯(x,y)) > b for all (x,y) ∈ P(ω,b,c). This means that the condition (i) of Theorem 3.7 is fulfilled.

Finally, we assume that (x,y)(t) ∈ 𝒫(ω,b,c) with ||𝒯 (x,y)|| > d, where b < dc. Then, we have bx(t) ≤ c and by(t) ≤ c for t ∈ [τ1, τ2]. From condition (H8) and Lemma 2.7, we get

ω ( T ( x , y ) ( t ) ) = inf τ 1 t τ 2 1 1 + ( log t ) p 1 + ( log t ) q 1 [ 1 G 1 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s + 1 G 2 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 G 3 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) d s s + 1 G 4 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) d s s ] 1 inf τ 1 t τ 2 G 1 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s + 1 inf τ 1 t τ 2 G 2 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s + 1 inf τ 1 t τ 2 G 3 ( t , s ) b ( s ) g ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s + 1 inf τ 1 t τ 2 G 4 ( t , s ) a ( s ) f ( x ( s ) , y ( s ) ) 1 + ( log t ) p 1 + ( log t ) q 1 d s s > b 2 Λ 3 n 3 ( m 1 + m 4 ) + b 2 Λ 4 n 4 ( m 2 + m 3 ) = b .

It follows that the condition (iii) of Theorem 3.7 is satisfied. Hence, by applying Theorem 3.7, we deduce that the problem (4) has at least three positive solutions (x1, y1), (x2, y2) and (x3, y3) such that ||(x1, y1|| < a, i n f τ 1 t τ 2 ( x 2 , y 2 ) ( t ) > b a n d ( x 3 , y 3 ) > a w i t h i n f τ 1 t τ 2 ( x 3 , y 3 ) ( t ) < b . This completes the proof.  □

4 Examples

In this section, we present two examples to illustrate our results.

Example 4.1

Consider the following Hadamard fractional differential system subject to boundary conditions on an unbounded domain

D3/2x(t)+erf(x(t),y(t))=0,t(1,),D5/3y(t)+t2g(x(t),y(t))=0,t(1,),x(1)=0,D1/2x()=23I1/2y(74)+π5I3/2y(74),y(1)=0,D2/3y()=25I1/3x(83)+17I2/3x(83)+2eI4/3x(83),(20)
where
f(x,y)=x1+y(14x)+y1+x(14y)+7;0x,y1/4,6+e116xy+sin2((14x)(14y));1/4x,y<,
and
g(x,y)={ x 2 1+ x 2 ( 1 4y)+ y 4 1+ y 4 ( 1 4x)+2;0x,y1/4, 2+exysin4(( x 1 4 )( y 1 4 ));1/4x,y<.

Here p=3/2,q=5/3,a(t)=et,b(t)=t2,m=2,η=7/4,λ1=2/3,α1=1/2,λ2=π/5,α2=3/2,n=3,ξ=8/3,σ1=2/5,β1=1/3,σ2=1/7,β2=2/3,σ3=2/eandβ3=4/3. We find that Λ1 = 0.3512388401 and Λ2 = 0.9782224108 which leads to Ω = 0.4564474804 > 0. In addition, we can compute that M1 = 1.977763776, M2 = 0.7695054857, M3 = 1.941574800, M4 = 2.143121504, m1 = 0.2825156640, m2 = 0.2371688476, m3 = 0.2332983559, m4 = 0.7105323246, n1 = 0.2193839344 and n2 = 0.5000000000. By choosing k1 = 2 and k2 = 3, we also obtain n3 = 0.01925300492, n4 = 0.1451247166, Λ3 = 0.01911915781, Λ4 = 0.06827641958, Λ5 = 0.9040560259 and Λ6 = 1.355540143. Observe that the functions f, g, a and b satisfy the conditions .(H1)-(H2).

Choosing ρ1 = 1/4, ρ2 = 100, θ1 = 54 ∈ (Λ-13, ∞) = (52.30355908, ∞), θ2 = 16 ∈ (Λ-14, ∞) = (14.64634505, ∞) θ3 = 1 ∈ (0, Λ-15) = (0, 1.106126137) and θ4 = (0.5 ∈ (0, Λ-16) = (0, 0.7377133058), we obtain

f(x,y)7θ1ρ12andg(x,y)2θ2ρ12,

for 0 ≤ x,y ≤ 1/4. Also we have

f(x,y)50θ3ρ22andg(x,y)25θ4ρ22,

for 1/4 ≤ x, y < ∞.

Hence the conditions (H3)-(H4) hold. By Theorem 3.4, we conclude that the problem (20) has at least one positive solution (x,y) such that 1/4 < ||(x,y)|| < 100.

Example 4.2

Consider the following Hadamard fractional differential system subject to boundary conditions on an unbounded domain

D7/4x(t)+t3/4f(x(t),y(t))=0,t(1,),D9/5y(t)+e2tg(x(t),y(t))=0,t(1,),x(1)=0,D3/4x()=π13I1/4y(95)+712I1/2y(95)+215I3/4y(95)y(1)=0,D4/5y()=316I1/5x(73)+25I2/5x(73)+13e2I3/5x(73)+38πI4/5X(73),(21)

where

f(x,y)=x5(1+x)(35x)+y5(1+y)(35y)+3100;0x,y3/5,3100e|xy|+3(x35)+2(y35);3/5x,y7/5,403100+e((7/5)y)sin2(75x);7/5x,y<,

and

g(x,y)=xy(35x)(35y)+2;0x,y3/5,2e|x2y2|+230(x35)+220(y35);3/5x,y7/5,362+e2((7/5)x)sin4(75y);7/5x,y<.

Here p=7/4,q=9/5,a(t)=t3/4,b(t)=e2t,m=3,η=9/5,λ1=π/13,α1=1/4,λ2=7/12,α2=1/2,λ3=2/15,α2=3/4,n=4,ξ=7/3,σ1=3/16,β1=1/5,σ2=2/5,β2=2/5,σ3=1/e2 and β3 = 3/5, σ4 = 3/8ᅲ and β4 = 4/5. We find that Λ1 = 0.3324581038 and Λ2 = 0.8725814055 which leads to Ω = 0.5659031627 > 0. In addition, we can compute that M1 = 1.645835954, M2 = 0.5874823216, M3 = 1.624063246, M4 = 1.541927070, m1 = 0.1785392532, m2 = 0.1696242483, m3 = 0.1547484152, m4 = 0.4894898091, n1 = 1.333333333 and n2 = 0.04890051071. By choosing k1 = 4 and k2 = 5, it is easy to see that [ξ, k1ξ] ∩ [η, k2η] ≠θ. Then we also obtain n3 = 0.4565504861, n4 = 0.006160413528, Λ3 =0.3049889931, Λ4 = 0.001998269744, Λ5 = 4.250350698 and Λ6 = 0.1081457077. From above information, the conditions (H1)-(H2) are fulfilled.

Choosing a = 3/5, b = 7/5, c = 80, we get

f(x,y)0.06600000000andg(x,y)2.008100000000,

which yields for 0 ≤ x, y ≤ 3/5,

f(x,y)<0.07058241103=a2Λ5andg(t,x,y)<2.774035201=a2Λ6.

In addition, we obtain

f(x,y)4.030000000andg(x,y)362.000000,

which leads to

f(x,y)>2.295164796=b2Λ3andg(x,y)>350.3030570=b2Λ4,

for 7/5 ≤ x, y ≤ 80. Also we have for 0 ≤ x, y ≤ 80.

f(x,y)9.410988138=c2Λ5andg(x,y)369.8713601=c2Λ6.

It is easy to see that τ1=max{ξ,η}=7/3,τ2=min{k1ξ,k2η}=9.

Therefore, the conditions (H7)-(H9) of Theorem 3.8 hold. Applying Theorem 3.8, we deduce that the problem (21) has at least three positive solutions (x1, y1), (x2, y2) and (x3, y3) such that ||(x1, y1|| < 3/5, inf7/3t9(x2,y2)(t)>7/5and(x3,y3)>3/5withinf7/3t9(x3,y3)(t)<7/5.

Acknowledgement

This research was funded by King Mongkut’s University of Technology North Bangkok. Contract no. KMUTNB-GEN-59-18.

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Received: 2016-10-10
Accepted: 2017-3-21
Published Online: 2017-5-25

© 2017 Tariboon et al.

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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  232. The algebraic size of the family of injective operators
  233. Topical Issue on Topological and Algebraic Genericity in Infinite Dimensional Spaces
  234. The history of a general criterium on spaceability
  235. Topical Issue on Topological and Algebraic Genericity in Infinite Dimensional Spaces
  236. On sequences not enjoying Schur’s property
  237. Topical Issue on Topological and Algebraic Genericity in Infinite Dimensional Spaces
  238. A hierarchy in the family of real surjective functions
  239. Topical Issue on Topological and Algebraic Genericity in Infinite Dimensional Spaces
  240. Dynamics of multivalued linear operators
  241. Topical Issue on Topological and Algebraic Genericity in Infinite Dimensional Spaces
  242. Linear dynamics of semigroups generated by differential operators
  243. Special Issue on Recent Developments in Differential Equations
  244. Isomorphism theorems for some parabolic initial-boundary value problems in Hörmander spaces
  245. Special Issue on Recent Developments in Differential Equations
  246. Determination of a diffusion coefficient in a quasilinear parabolic equation
  247. Special Issue on Recent Developments in Differential Equations
  248. Homogeneous two-point problem for PDE of the second order in time variable and infinite order in spatial variables
  249. Special Issue on Recent Developments in Differential Equations
  250. A nonlinear plate control without linearization
  251. Special Issue on Recent Developments in Differential Equations
  252. Reduction of a Schwartz-type boundary value problem for biharmonic monogenic functions to Fredholm integral equations
  253. Special Issue on Recent Developments in Differential Equations
  254. Inverse problem for a physiologically structured population model with variable-effort harvesting
  255. Special Issue on Recent Developments in Differential Equations
  256. Existence of solutions for delay evolution equations with nonlocal conditions
  257. Special Issue on Recent Developments in Differential Equations
  258. Comments on behaviour of solutions of elliptic quasi-linear problems in a neighbourhood of boundary singularities
  259. Special Issue on Recent Developments in Differential Equations
  260. Coupled fixed point theorems in complete metric spaces endowed with a directed graph and application
  261. Special Issue on Recent Developments in Differential Equations
  262. Existence of entropy solutions for nonlinear elliptic degenerate anisotropic equations
  263. Special Issue on Recent Developments in Differential Equations
  264. Integro-differential systems with variable exponents of nonlinearity
  265. Special Issue on Recent Developments in Differential Equations
  266. Elliptic operators on refined Sobolev scales on vector bundles
  267. Special Issue on Recent Developments in Differential Equations
  268. Multiplicity solutions of a class fractional Schrödinger equations
  269. Special Issue on Recent Developments in Differential Equations
  270. Determining of right-hand side of higher order ultraparabolic equation
  271. Special Issue on Recent Developments in Differential Equations
  272. Asymptotic approximation for the solution to a semi-linear elliptic problem in a thin aneurysm-type domain
  273. Topical Issue on Metaheuristics - Methods and Applications
  274. Learnheuristics: hybridizing metaheuristics with machine learning for optimization with dynamic inputs
  275. Topical Issue on Metaheuristics - Methods and Applications
  276. Nature–inspired metaheuristic algorithms to find near–OGR sequences for WDM channel allocation and their performance comparison
  277. Topical Issue on Cyber-security Mathematics
  278. Monomial codes seen as invariant subspaces
  279. Topical Issue on Cyber-security Mathematics
  280. Expert knowledge and data analysis for detecting advanced persistent threats
  281. Topical Issue on Cyber-security Mathematics
  282. Feedback equivalence of convolutional codes over finite rings
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